Calculate the power used in the $ 2\Omega $ resistor in each of the following circuits: (i) a $ 6V $ battery in series with $ 1\Omega $ and $ 2\Omega $ resistors, and (ii) a $ 4V $ battery in parallel with $ 1\Omega $ and $ 2\Omega $ resistors.

Hint
We have to calculate the power consumed by a resistor in case of a series and parallel connection. We can calculate the power with the help of an expression that tells us that power is equal to the square of the voltage across the resistor divided by the value of the resistance.
$\Rightarrow {{\text{R}}_{\text{series}}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+--+{{R}_{n}} $ , $ P=\dfrac{{{V}^{2}}}{R} $ , $ {{V}_{R}}=\dfrac{R}{{{\operatorname{R}}_{eq}}}\times V $

Complete step by step answer
We have been given two cases. In the first situation, we have a battery and two resistances connected in series with it.
The voltage across the resistor can be found using the voltage divider rule which states that in a series connection, the voltage across a resistance $ R $ can be given as $ \dfrac{R}{{{\operatorname{R}}_{eq}}}\times V $ where $ {{\operatorname{R}}_{eq}} $ is the equivalent resistance of the circuit and $ V $ is the voltage source connected in the circuit.
The equivalent resistance of a series circuit can be given as $ {{\text{R}}_{\text{series}}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+--+{{R}_{n}} $
Substituting the values for the given circuit, we can say that the equivalent resistance $ ({{\operatorname{R}}_{eq}})=(1+2)\Omega =3\Omega $
Hence the voltage across the $ 2\Omega $ resistor can be given as $ (V)=\dfrac{2\Omega }{3\Omega }\times 6V=4V $
Now the power consumed by the resistor can be given as $ P=\dfrac{{{V}^{2}}}{R} $
Substituting the values, we get $ (P)=\dfrac{{{\left( 4V \right)}^{2}}}{2\Omega }=8W $
Moving on to the second scenario, the resistors are connected in parallel across a $ 4V $ battery
In a parallel combination, the voltage remains the same across every resistor
Hence voltage across the $ 2\Omega $ resistor will be $ 4V $
The power can be calculated as $ ({{P}_{2}})=\dfrac{{{\left( 4V \right)}^{2}}}{2\Omega }=8W $
Hence we can say that the power consumed in both the given cases would be $ 8W $ .

Note
In a series circuit, voltage division takes place across each resistor whereas, in a parallel combination, current division across each resistor takes place. The rules for current division and voltage divisions should be remembered as the values of the resistor that should be applied often creates confusion. In voltage division, the value of the resistor itself is used but in the current division, the value of the adjacent resistor is used.